Maths Olympiad Prep

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Algebra Difficulty 5.2 AIME, harder Find the answer

We can view these conditions as a geometry diagram as seen below. So, we know that ef=34\frac{e}{f}=\frac{3}{4} (since e=ab=34c34d=34fe=a-b=\frac{3}{4} c-\frac{3}{4} d=\frac{3}{4} f and we know that e2+f2=15\sqrt{e^{2}+f^{2}}=15 (since this is a2+c2b2+d2)\left.\sqrt{a^{2}+c^{2}}-\sqrt{b^{2}+d^{2}}\right). Also, note that ac+bdadbc=(ab)(cd)=efa c+b d-a d-b c=(a-b)(c-d)=e f. So, solving for ee and ff, we find that e2+f2=225e^{2}+f^{2}=225, so 16e2+16f2=360016 e^{2}+16 f^{2}=3600, so (4e)2+(4f)2=3600(4 e)^{2}+(4 f)^{2}=3600, so (3f)2+(4f)2=3600(3 f)^{2}+(4 f)^{2}=3600, so f2(32+42)=3600f^{2}\left(3^{2}+4^{2}\right)=3600, so 25f2=360025 f^{2}=3600, so f2=144f^{2}=144 and f=12f=12. Thus, e=3412=9e=\frac{3}{4} 12=9. Therefore, \boldsymbol{e f}=\mathbf{9} * \mathbf{1 2}=\mathbf{1 0 8}$.

A number or a short expression. Spacing and $ signs are ignored.

Solution

The value of efef is 108.

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.